A solenoid 0.70 m in length and of 2100 turns has a radius `4.5xx10^(-2) `m. A second coil of 750 turns is wound over the middle part of the solenoid. Find the self inductance of the solenoid and mutual inductance between the two coils.

Explanation

Given , l = 0.70 m. `N_1= 2100 , r=4.5 xx10^(-2)m N_2=750 , A = pir^2`.
Self inductance of the solenoid, `L= (mu_0AN_1^2)/l` for air core.
i.e., `L= (4xx3.142xx10^(-7)xx3.142xx(4.50xx10^(-2))^2xx(2100)^2)/(0.70) = 52.3xx10^(-3)H`
Mutual inductance M `= (mu_0N_1N_2A)/l=(4xx3.142xx10^(-7)xx2100xx750xx3.142xx(4.5xx10^(-2))^2)/(0.70)=18xx10^(-3)H`